Հանրահաշիվ

Ա) a_n = (2n/n+1/n)/(5n/n+3/n) = (2+1/n)/(5-3/n) = (2+0)/(5-0) = 2/5

Բ) a_n = (4n/n-5/n)/(8n/n+3/n) = (4-5/n)/(8+3/n) = (4-0)/(8+0) = 4/8 = 1/2

Գ) a_n = (5n/n-√n/n-3/n)/(n/n+2√n/n+4/n) = (5-√n/n-3/n)/(1+0+4/n) = (5-0-0)/(1+0+0) = 5/1 = 5

Դ) a_n = (3n/n+5^3√n-8/n)/(2n/n-3√n/n+9/n) = (3+0-0)/(2-0+0) = 3/2

Ա) limn>∞ (4n^3+3n^2-200)/(2x^3-2n+12) = limn>∞(4n^3/n^3+3n^2/n^3-200/n^3)/(2n^3/n^3-2n/n^3+12/n^3) = limn>∞ (4+0-0)/(2-0-0) = 4/2 = 2

Բ) limn>∞ (5n^4-1)/(n-2n^4) = limn>∞(5n^4/n^4-1/n^4)/(n/n^4-2n^4/n^4) = limn>∞ (5-1/n^4)/(1/n^3-2) = limn>∞ (5-0)/(0-2) = 5/-2 = (-5/2)

Գ) limn>∞ (2n^99-n^21)/(2n^21-4n^99+1) = limn>∞ (2n^99/n^99-n^21/n^99)/(2n^21/n^99-4n^99/n^99+1/n^99) = limn>∞(2-0)/(0-4+0) = 2/-4 = -(2/4)

Դ) limn>∞ (7n^5-1)/(n^5-n^3+1) = limn>∞ (7n^5/n^5-1/n^5)/(n^5/n^5-n^3/n^5+1/n^5) = limn>∞ (7-0)/(1-0+0) = 7/1 = 7

Ա) (n/n-1/n)/(1/n+n^2/n) = (1-0)/(0+n) = 1/n = 0

Բ) ((n^12)-(n^11))/((n^11)-(2(n^13))) = limn>∞(n^11(n-1))/(n^11(1-2n^2)) = limn>∞(n-1)/(1-2n^2) = limn>∞ 1/-4n = 0

Գ) (1/n-n^3/n+n/n)/(n^2/n+n^5/n) = ((1/n)-((n^3)/n)+(n/n))/(((n^2)/n)+((n^5)n))) = (1/n+n^2+1)/(n+n^5) = (0-n+1)/(n+n) = limn>∞ (-2n)/(1+5n^4) = limn>∞ (-2n)/n = 0

Ա) lim(n→∞)⁡(√(n+100)-√n)=(√(n+100)-√n)(√(n+100)+√n)/(√(n+100)+√n)=(√(n+100)√(n+100)+√(n+100)√n-√n√(n+100)-√n√n)/(√(n+100)+√n)=(√((n+100)^2 )+√(n+100)√n-√n√(n+100)-(√n)^2)/(√(n+100)+√n)=((√(n+100))^2-)√n)^2)/(√(n+100)+√n)=lim(n→∞)⁡(100/(√(n+100)+√n)= 0

Բ) lim(n→∞)⁡(√(n^2+1)-n)=lim(n→∞)⁡((√(n^2+1)-n)(√(n^2+1)+n)/((√(n^2+1)+n) ))=((√(n^2+1))^2-n^2)/(√(n^2+1)+n)=lim(n→∞)⁡(n^2+1)^ -n^2)/(√(n^2+1)+n))=lim(n→∞)⁡(1/(√(n^2+1)+n))=0

Գ) lim(n→∞)⁡((√(n+2)-√(n+1))/(√(n+1)-√n))=lim┬(n→∞)⁡(((√(n+2)-√(n+1))/(√(n+1)-√n)*(√(n+2)+√(n+1))/(√(n+1)+√n))/((√(n+2)+√(n+1))/(√(n+1)+√n)))=((√(n+2)-√(n+1))(√(n+2)+√(n+1))/(√(n+1)-√n)(√(n+1)+√n) )/((√(n+2)+√(n+1))/(√(n+1)+√n))=((√((n+2)^2 )-(√(n+1))^2)/(√((n+1)^2 )-(√n)^2 ))/((√(n+2)+√(n+1))/(√(n+1)+√n))=1/((√(n+2)+√(n+1))/(√(n+1)+√n))=lim(n→∞)⁡((1/(1+√((n+1)/(n+2))))/(√(n+1)/√(n+2)+√n/√(n+2)))=lim(n→∞)⁡(1/(((1+√((n+1)/(n+2)))/(√(n+1)/√(n+2)+√n/√(n+2))) ))=1/(((1+1)/(1+1)) )=1/((2/2) )=1

Դ) lim(n→∞)⁡((√(n+1)-√n) √(n-1))=0*lim(n→∞)⁡√(n-1)=0